Divide the following complex numbers. $\dfrac{-14+22i}{-2-4i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate, which is ${-2+4i}$. $ \dfrac{-14+22i}{-2-4i} = \dfrac{-14+22i}{-2-4i} \cdot \dfrac{{-2+4i}}{{-2+4i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$. $ = \dfrac{(-14+22i) \cdot (-2+4i)} {(-2)^2 - (-4i)^2} $ Evaluate the squares in the denominator and subtract them. $ = \dfrac{(-14+22i) \cdot (-2+4i)} {(-2)^2 - (-4i)^2} $ $ = \dfrac{(-14+22i) \cdot (-2+4i)} {4 + 16} $ $ = \dfrac{(-14+22i) \cdot (-2+4i)} {20} $ The denominator now doesn't contain any imaginary unit multiples, so it is a real number. Note that when a complex number, $a + bi$ is multiplied by its conjugate, the product is always $a^2 + b^2$. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-14+22i}) \cdot ({-2+4i})} {20} $ $ = \dfrac{{-14} \cdot {(-2)} + {22} \cdot {(-2) i} + {-14} \cdot {4 i} + {22} \cdot {4 i^2}} {20} $ $ = \dfrac{28 - 44i - 56i + 88 i^2} {20} $ Finally, simplify the fraction. $ \dfrac{28 - 44i - 56i - 88} {20} = \dfrac{-60 - 100i} {20} = -3-5i $